Question

Consider the vector field.f(x, y, z) = xy2z2i + x2yz2j + x2y2zk(a) find the curl of the vector field.

Answer 1

First introduce the following notations:
$P=yx^2z^2\\Q=x^2yz^2\\R=x^2y^2z\\\text{Then compute the partial derivative like this:}\\R_y=2x^2yz\\Q_z=2x^2yz\\P_z=2yx^2z\\R_x=2xy^2z\\Q_x=2xyz^2\\P_y=x^2z^2\\\text{Then evaluate each of the following:}\\R_y-Q_z=2x^2yz-2x^2yz=0\\P_z-R_x=2yx^2z-2xy^2z=2xyx(x-y)\\Q_x-R_y=2xyz^2-x^2z^2=z^2x(2y-x)\\\text{Our answer will be then:}\\Curl(f)=(2xyx(x-y))j+(z^2x(2y-x))k$